jas*_*son 25 ajax jquery multipartform-data apache-commons-fileupload
我有一个HTML表单,需要在一个请求中将3个部分上传到现有的REST API.我似乎无法找到有关如何在FormData提交上设置边界的文档.
我试图按照这里给出的示例: 如何在jQuery中使用Ajax请求发送FormData对象?
但是,当我提交数据时,它会被以下堆栈跟踪拒绝:
Caused by: org.apache.commons.fileupload.FileUploadException: the request was rejected because no multipart boundary was found.
我该如何设置边界?
这是HTML/Javascript:
   <script type="text/javascript">
    function handleSubmit() {
        var jsonString = "{" +
                "\"userId\":\"" + document.formSubmit.userId.value + "\"" +
                ",\"locale\":\"" + document.formSubmit.locale.value + "\"" +
                "}";
        var data = new FormData();
        data.append('Json',jsonString);
        data.append('frontImage', document.formSubmit.frontImage.files[0]);
        data.append('backImage', document.formSubmit.backImage.files[0]);
        document.getElementById("sent").innerHTML = jsonString;
        document.getElementById("results").innerHTML = "";
        $.ajax({
                   url:getFileSubmitUrl(),
                   data:data,
                   cache: false,
                   processData: false,
                   contentType: 'multipart/form-data',
                   type:'POST',
                   success:function (data, status, req) {
                       handleResults(req);
                   },
                   error:function (req, status, error) {
                       handleResults(req);
                   }
               });
    }
</script>
这是表格:
<form name="formSubmit" action="#">
    userId: <input id="userId" name="userId" value=""/><br/>
    locale: <input name="locale" value="en_US"/><br/>
    front Image: <input type="file" name="frontImage"/><br/>
    back Image: <input type="file" name="backImage"/><br/>
    <input type="button" onclick="handleSubmit();" value="Submit"/>
</form>
在此先感谢您的帮助!
jas*_*son 22
穆萨的反应非常好.将contentType设置为false确实正确提交了表单数据.谢谢!
这是一个有效的ajax调用:
$.ajax({
    url:getFileSubmitUrl(),
    data:data,
    cache:false,
    processData:false,
    contentType:false,
    type:'POST',
    success:function (data, status, req) {
        handleResults(req);
    },
    error:function (req, status, error) {
        handleResults(req);
    }
});
我还发现这段代码也有效:
  var oReq = new XMLHttpRequest();
        oReq.open("POST", getFileSubmitUrl());
        oReq.addEventListener("error", transferComplete);
        oReq.addEventListener("load", transferComplete);
        oReq.addEventListener("abort", transferComplete);
        oReq.send(data);
    }
    function transferComplete(evt) {
        handleResults(evt.target);
    }
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